Đáp án:
$a,S=\{-20\}$
$b,S=\left\{\dfrac{9}{4}\right\}$
$c,S=\{9\}$
Giải thích các bước giải:
a) ĐKXĐ: $x\le \dfrac{4}{3}$
$\sqrt{4-3x}=8$
$⇔4-3x=64$
$⇔3x=-60$
$⇔x=-20\,(TM)$
Vậy $S=\{-20\}$.
b) ĐKXĐ: $x\ge 2$
$\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=-1$
$⇔2\sqrt{x-2}-4\sqrt{x-2}=-1$
$⇔-2\sqrt{x-2}=-1$
$⇔\sqrt{x-2}=\dfrac{1}{2}$
$⇔x-2=\dfrac{1}{4}$
$⇔x=\dfrac{9}{4}\,(TM)$
Vậy $S=\left\{\dfrac{9}{4}\right\}$.
c) ĐKXĐ: $x\ge 0$
$(2\sqrt x+1)(\sqrt x-2)=7$
$⇔2x-4\sqrt x+\sqrt x-2=7$
$⇔2x-3\sqrt x-9=0$
$⇔(2\sqrt x+3)(\sqrt x-3)=0$
Vì $2\sqrt x+3>0$
$⇒\sqrt x-3=0$
$⇒\sqrt x=3$
$⇒x=9\,(TM)$
Vậy $S=\{9\}$.
